One other way will be using two pointers.
step1) Take two pointers(p1 and p2 ), pointing to the beginning of
list L1 and L2.
step2) Now start moving both the pointers simultaneously and check
whether they point to same node. If not, then move both pointers to
next node.
Step3) If any of pointer becomes NULL then set it to the beginning of
another list. Lets say p1 becomes NULL, then set p1 to point to L2.
Step4) Repeat the step2
After two traversals, both pointers will surely meet at intersection
point.

On Oct 10, 3:24 pm, Manisha <pgo...@gmail.com> wrote:
> @sandeep
> notice that solution given on the given link doesn't satisfy the
> conditions given in the above question.
>
> @sharad
> Both lists may have duplicate values. So in this case it will better
> to hash the address of node instead of values.
>
> One other way is to maintain a flag in each of the node. In the
> begining flag will be 0 for all the items in both list. While
> traversing the first list, set the flags to 1. While traversing the
> seconds list,look for the item for which flag has been set.  But this
> will also take additional o(n) space.
>
> I was thinking in line of reversing the first list while traversing it
> (it will use recursion). However, I couldn't make it work. Any
> thoughts?
>
> On Oct 10, 3:02 am, sandeep jain <jainsandee...@gmail.com> wrote:
>
>
>
>
>
> > Here is one solutionhttp://geeksforgeeks.org/?p=2405
>
> > On Fri, Oct 9, 2009 at 9:00 AM, sharad kumar <aryansmit3...@gmail.com>wrote:
>
> > > space comp O(n)
> > > time o(2n) both in terms of  worst case
>
> > > On Fri, Oct 9, 2009 at 8:46 PM, ankur aggarwal 
> > > <ankur.mast....@gmail.com>wrote:
>
> > >> @sharad
>
> > >> wat about space ??
> > >> extra space ?

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